Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 10 - Quadratic Equations - Chapter 10 Test - Page 469: 11



Work Step by Step

Using the rules for factoring trinomials, we obtain: $y^{2}+10y=24$ $y^{2}+10y-24=0$ $y^{2}-2y+12y-24=0$ $y(y-2)+12(y-2)=0$ $(y-2)(y+12)=0$ $(y-2)=0$ or $(y+12)=0$ $y=2$ or $y=-12$ Therefore, the solution set is {$-12,2$}.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.